Can superconductivity be understood to be a result of quantum entanglement? I've been trying to find out what causes superconductivity. I have seen a lot about what characterizes it (e.g the Meissner effect). But I haven't found much on the root cause.
The closest I have found to an actual root cause are "Cooper electron pairs", but even this is unclear to me.
I have a suspicion that quantum entanglement is what gives rise to superconductors and that Cooper pairs are entangled electrons. Is this suspicion accurate or am I off base? 
 A: The BCS state describing a conventional superconductor is indeed an entangled state, involving a superposition of different numbers of Cooper pairs, each of which involves a superposition of different combinations of paired momenta. This is a loose verbal description of equation (40) in

*

*Bardeen, Cooper, & Schrieffer, "Chapter 10: Superconductivity" (2017),  http://www.phys.lsu.edu/~jarrell/COURSES/SOLID_STATE/Chap10/chap10.pdf
However, although this kind of entanglement may be a necessary condition for  superconductivity, entanglement itself is certainly not a sufficient condition. The electrons in an ordinary non-superconducting metal are also in an entangled state, so I don't think we can say that entanglement "causes" superconductivity.
Entanglement is ubiquitous — it is the norm rather than the exception. Unentangled states (states in which each particle has a wavefunction of its own) are the exceptional ones.
An important thing about the BCS state is the specific way in which the electrons are entangled with each other. BCS superconductivity relies on the fact that many Cooper pairs can occupy the same "state," using the word "state" here like the word "orbital" in atomic physics. The particular way in which the electrons are entangled with each other is what makes this possible despite the fact that electrons individually obey the Pauli exclusion principle.
Here's the rough idea: let $|0\rangle$ denote the ground state and let $c^\dagger(k,s)$ denote the operator that promotes a single electron with spin $s=\pm 1/2$ (up or down) to an excited state with momentum $k$. Then an operator of the form
$$
    A \equiv \sum_{k,s} A(k,s)c^\dagger(k,s),
$$
with complex coefficients $A(k,s)$ creates a single electron with some generic wavefunction. Electrons are fermions, which means that the operators $c^\dagger(k,s)$ all anticommute with each other.
(In particular, any such operator multiplied by itself gives zero — this  is the Pauli exclusion principle.) So we can't promote two electrons into the same "state" (again in the sense of "orbital"), because
\begin{align*}
   A^2|0\rangle 
&=
 \left(\sum_{k,s} A(k,s)c^\dagger(k,s)\right)
 \left(\sum_{k',s'} A(k',s')c^\dagger(k',s')\right)|0\rangle \\
&=
 \sum_{k,s}\sum_{k',s'}
  A(k,s)
  A(k',s')c^\dagger(k,s)c^\dagger(k',s')|0\rangle \\
&=
 \sum_{k,s}\sum_{k',s'}
  A(k,s)
  A(k',s')\frac{c^\dagger(k,s)c^\dagger(k',s')+c^\dagger(k',s')c^\dagger(k,s)}{2}|0\rangle \\
&= 0
\end{align*}
where the second-to-last step follows because the sums are invariant under the exchange of $(k,s)$ and $(k',s')$. The zero state-vector does not represent any physical state, so two electrons cannot occupy the same "state." (This sentence uses the word "state" with two different meanings: the first one means the overall state of the system, and the second one is like "orbital." If I had the power to revise the established langauge, I would!) But now consider the operator
$$
    B \equiv \sum_{k,s} B(k,s)c^\dagger(k,s)c^\dagger(-k,-s),
$$
which is a rough analogue of the operator that creates a Cooper pair. The momenta and spins of the two electrons in this pair are entangled with each other (because the sum cannot be factorized). As a result of this entanglement, we have
$$
   B^n|0\rangle\neq 0
$$
even for large $n\gg 1$. The individual electrons still obey the Pauli exclusion principle (the operators $c^\dagger$ still anticommute with each other), but their entanglement means that plenty of terms in the product still survive even after taking this anticommutativity into account, leaving a nonzero result.
By the way, this is also why some atoms can behave as bosons (which don't obey the Pauli exclusion principle), despite being made of fermions (which do). So, again, entanglement is a necessary condition for conventional superconductivity, but not a sufficient condition.
